2 complex exercises I'm having trouble with

Question

$x+iy=\frac{z_1+z_2}{z_1-z_2}\qquad x,y\in\mathbb R$

Find $x, y$. And this one:

$z=(\sqrt{2+\sqrt2}+i\sqrt{2-\sqrt2})^4\implies|z|=\,?$

For the first one, I want to learn how to approach this kind of problem where I have $z_1$, $z_2$ and to find something about them, that doesn't work written as $z = a + bi$.

And for the second one, I tried first raising to the second power and then again to the second power but didn't got anywhere near the right answer and then tried to write it with in the trigonometric form but find that $\phi = \arctan(\sqrt2-1)$ which is not really helpful.

Answer 1

I suppose $z_1$ and $z_2$ are given complex numbers. To find $x$ and $y$ write $\frac {z_1+z_2} {z_1-z_2}$ as $\frac {(z_1+z_2)(\bar {z}_1-\bar z_2)} {(z_1-z_2)((\bar {z}_1-\bar z_2))}$. Note that the denominator is $|z_1-z_2|^{2}$ which is a positive number. Now you should be able to simplify the numerator and get the real and imaginary. For the second question use the fact that $|c^{4}|=|c|^{4}$. Hence $|z|=\{2+\sqrt 2 +2 -\sqrt 2\}^{2}=16$.

Answer 2

For the second question $$z=(\sqrt{2+\sqrt2}+i\sqrt{2-\sqrt2})^4\implies$$

$$|z|= |(\sqrt{2+\sqrt2}+i\sqrt{2-\sqrt2})|^4\implies$$

$$|z|= ({(\sqrt{2+\sqrt2})^2 +(\sqrt{2-\sqrt2})^2})^2=16$$

For the first question

$$ x+iy=\frac{z_1+z_2}{z_1-z_2}\qquad x,y\in\mathbb R$$

Let $ z_1=x_1+iy_1$ and $ z_2=x_2+iy_2$

$$\frac{z_1+z_2}{z_1-z_2}\qquad = \frac{(x_1+x_2)+i(y_1+y_2)}{(x_1+x_2)-i(y_1+y_2)}\qquad =x+iy $$

Where$$ x= \frac {(x_1+x_2)^2-(y_1+y_2)^2}{(x_1+x_2)^2+(y_1+y_2)^2}$$

and $$y= \frac {2(x_1+x_2)(y_1+y_2)}{(x_1+x_2)^2+(y_1+y_2)^2}$$